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9.04 Volume of spheres and hemispheres


There are many applications of finding the volume of a sphere. Below are just a few examples:

  • A toy manufacturer needs to figure out how much plastic they need to make super bouncy rubber balls would need to know the volume of the ball
  • An astronomer who wants to calculate how much the sun weighs would need to know its volume
  • A pharmaceutical company making round pills needs to know the dosage of medicine in each pill, which is found by its volume
  • A party balloon gas hire company needs to know how much gas fits inside each of their balloons, which is found by its volume


The formula

The volume of a sphere with radius $r$r can be calculated using the following formula:

Volume of sphere

$\text{Volume of sphere }=\frac{4}{3}\pi r^3$Volume of sphere =43πr3


Proof of the formula

We won't be expected to recreate this proof, but it can be interesting to read through.

If four points on the surface of a sphere are joined to the center of the sphere, then a pyramid of perpendicular height r is formed, as shown in the diagram. Consider the solid sphere to be built with a large number of these solid pyramids that have a very small base which represents a small portion of the surface area of a sphere.

If $A_1$A1, $A_2$A2, $A_3$A3, $A_4$A4, .... , $A_n$An represent the base areas (of all the pyramids) on the surface of a sphere (and these bases completely cover the surface area of the sphere), then,

$\text{Volume of sphere }$Volume of sphere $=$= $\text{Sum of all the volumes of all the pyramids }$Sum of all the volumes of all the pyramids
$V$V  $=$= $\frac{1}{3}A_1r+\frac{1}{3}A_2r+\frac{1}{3}A_3r+\frac{1}{3}A_4r$13A1r+13A2r+13A3r+13A4r   $\text{+ ... +}$+ ... + $\frac{1}{3}A_nr$13Anr
  $=$= $\frac{1}{3}$13  $($(   $A_1+A_2+A_3+A_4$A1+A2+A3+A4  $\text{+ ... +}$+ ... +  $A_n$An  $)$) $r$r 
  $=$= $\frac{1}{3}\left(\text{Surface area of the sphere }\right)r$13(Surface area of the sphere )r
  $=$= $\frac{1}{3}\times4\pi r^2\times r$13×4πr2×r
  $=$= $\frac{4}{3}\pi r^3$43πr3

where $r$r is the radius of the sphere.  


Worked example

Question 1

Find the volume of a marble with a diameter of $2.3$2.3 cm, to two decimal places.

Think: We have been given the diameter instead of the radius, but the formula uses the radius. Since $r=\frac{D}{2}$r=D2, we can determine that the radius is $\frac{2.3}{2}$2.32 or $1.15$1.15 cm.


$V$V $=$= $\frac{4}{3}\pi r^3$43πr3

State the formula

$V$V $=$= $\frac{4}{3}\pi\times1.15^3$43π×1.153

Fill in the given information

$V$V $=$= $6.370626$6.370626...

Using a calculator, evaluate using the $\pi$π button

$V$V $=$= $6.37$6.37

Round to two decimals


Practice questions

Question 2

Find the volume of the sphere shown.

Round your answer to two decimal places.

Question 3

A sphere has a radius $r$r cm long and a volume of $\frac{343\pi}{3}$343π3 cm3. Find the radius of the sphere.

Round your answer to two decimal places.

Enter each line of working as an equation.



Use surface area and volume of three-dimensional objects to solve practical problems

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